# Deriviation of WKB approximation

• Repetit
In summary: No, it's not accurate. The approximation obtained by keeping a finite number of terms is not an asymptotic series. It's an exact series.
Repetit
Hey!

In deriving the WKB approximation the wave function is written as

$$\psi \left( x \right) = exp\left[ i S\left( x \right) \right ]$$

Now, in some of the deriviations I've seen, the function S(x) is expanded as a power series in $$\hbar$$ as

$$S(x) = S_0(x) + \hbar S_1(x) + \frac{\hbar}{2} S_2(x) ...$$

I don't really understand this. It's something like $$S_0$$ being the classical result and, the next term being a first order quantum correction and so on. But why do you choose to expand in powers of $$\hbar$$? Can somebody explain to me what this is all about?

René

That particular form for $S(x)$ has the correspondence principle built right into it. If you take the limit as $\hbar \rightarrow 0$, you recover the classical result.

Tom Mattson said:
That particular form for $S(x)$ has the correspondence principle built right into it. If you take the limit as $\hbar \rightarrow 0$, you recover the classical result.

(This thread appeared on Google and I have the exact same question) I am extremely confused at your statement. $\hbar$ is a constant, right? How on Earth can one construct a power series of a function S(x) by expanding it as a function of a constant? What does that even mean?

I have taken a few (more like 1.5 and some self study) classes in QM on the engineering side, but this is over my head. I've currently borrowed a few different QM textbooks and they all say the same opaque thing.

Last edited:
You'll be seeing a lot more constants being treated like parameters in physics, so you'll have to get used to it.

Lets parameterize all the possible universes by different values of $$\hbar$$, and solve quantum mechanics in each of them. Then you'd get a family of solutions parameterized by $$\hbar$$. If you choose our universe, corresponding to our $$\hbar$$, then in principle you have the solution to QM in our universe, no?

There's a caveat, of course. You have to assume that the solutions to problems behave smoothly with $$\hbar$$, which is a reasonable assumption, but only comes from experience.

Anyway, if you stick around long enough you'll get to differentiate with respect to orbital angular momentum $$\ell$$ and all sorts of goodness (Feynman-Hellman theorem)

I think I understand a little better now, but I'll try to explain what is bugging me still. After reading around, I've come to the conclusion a power series with respect to constants is not so far-fetched: For example, any decimal number can be expressed as a power series in 10, or any other number really.

However, the textbook I am primarily using ( Bransden & Joachain Quantum Mechanics: Second Edition ) mentions the power series for S(x) "does not converge, but is an asymptotic series for the function S(x). As a result, the best approximation to S(x) is obtained by keeping a finite number of terms". I've been reading about asymptotic series, but their rationale/use isn't very clear to me still.

I guess the closest I can come to what an asymptotic series means in words, it's "Adding more terms to the expansion won't make the relative error appreciably smaller". Is this accurate?

## 1. What is the WKB approximation?

The WKB (Wentzel-Kramers-Brillouin) approximation is a mathematical method used to approximate the solutions to certain differential equations, specifically those with rapidly varying coefficients. It is often used in quantum mechanics to approximate the wavefunction of a particle in a potential.

## 2. How is the WKB approximation derived?

The WKB approximation is derived by assuming that the wavefunction can be written as a product of two terms, one slowly varying and one rapidly varying. This allows the differential equation to be separated into two equations, with the rapidly varying term being treated as a perturbation. The approximation is then obtained by solving the perturbation equation.

## 3. What are the limitations of the WKB approximation?

The WKB approximation is only valid for certain types of differential equations, specifically those with rapidly varying coefficients. It also breaks down when the potential becomes too steep or when the energy of the particle becomes too low.

## 4. Can the WKB approximation be used for any potential?

No, the WKB approximation can only be used for potentials that are well-behaved and do not have sharp discontinuities. It also requires the potential to be slowly varying compared to the wavelength of the particle.

## 5. What is the significance of the WKB approximation in quantum mechanics?

The WKB approximation is a useful tool in quantum mechanics as it allows for the approximate calculation of wavefunctions and energies for certain systems. It also provides insight into the behavior of quantum systems and can be used to understand phenomena such as tunneling and bound states.

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